User Guide

Page 18-47
Inferences concerning one variance
The null hypothesis to be tested is , H
o
: σ
2
= σ
o
2
, at a level of confidence (1-
α)100%, or significance level α, using a sample of size n, and variance s
2
.
The test statistic to be used is a chi-squared test statistic defined as
2
0
2
2
)1(
σ
χ
sn
o
−
=
Depending on the alternative hypothesis chosen, the P-value is calculated as
follows:
• H
1
: σ
2
< σ
o
2
, P-value = P(χ
2
<χ
o
2
) = 1-UTPC(ν,χ
o
2
)
• H
1
: σ
2
> σ
o
2
, P-value = P(χ
2
>χ
o
2
) = UTPC(ν,χ
o
2
)
• H
1
: σ
2
≠ σ
o
2
, P-value =2⋅min[P(χ
2
<χ
o
2
), P(χ
2
>χ
o
2
)] =
2⋅min[1-UTPC(ν,χ
o
2
), UTPC(ν,χ
o
2
)]
where the function min[x,y] produces the minimum value of x or y (similarly,
max[x,y] produces the maximum value of x or y). UTPC(ν,x) represents the
calculator’s upper-tail probabilities for ν = n - 1 degrees of freedom.
The test criteria are the same as in hypothesis testing of means, namely,
• Reject H
o
if P-value < α
• Do not reject H
o
if P-value > α.
Please notice that this procedure is valid only if the population from which the
sample was taken is a Normal population.
Example 1
-- Consider the case in which σ
o
2
= 25, α=0.05, n = 25, and s
2
=
20, and the sample was drawn from a normal population. To test the
hypothesis, H
o
: σ
2
= σ
o
2
, against H
1
: σ
2
< σ
o
2
, we first calculate
2.189
25
20)125()1(
2
0
2
2
=
⋅−
=
−
=
σ
χ
sn
o
With ν = n - 1 = 25 - 1 = 24 degrees of freedom, we calculate the P-value as,
P-value = P(χ
2
<19.2) = 1-UTPC(24,19.2) = 0.2587…
Since, 0.2587… > 0.05, i.e., P-value > α, we cannot reject the null
hypothesis, H
o
: σ
2
=25(= σ
o
2
).